polygon triangulation proof

/PaintType 0 def 22 0 obj 1 Introduction 1.1 De nitions: The graph of triangulations 1.An n-gon is a regular polygon with n sides. We first establish a preliminary result: Every triangulation of an n-gon has (n-2)-triangles formed by (n-3) diagonals. We will have to wait for the discussion of triangulation to formally prove that triangulation graphs of polygons … The proof still holds even if we turned the polygon upside down. /FontFile 23 0 R diagonal splits P into polygons P 1 (m 1 vertices) and P 2 (m 2 vertices) both m 1 and m 2 must be less than n, so by induction P 1 and P 2 can be triangulated; hence, P can be triangulated << << endobj n 3 n 3 Proof: For the upper bound, 3-color any triangulation of the polygon and take the color with the minimum number of … 638.9 638.9 958.3 958.3 319.4 351.4 575 575 575 575 575 869.4 511.1 597.2 830.6 894.4 500 500 611.1 500 277.8 833.3 750 833.3 416.7 666.7 666.7 777.8 777.8 444.4 444.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 627.2 817.8 766.7 692.2 664.4 743.3 715.6 The triangulation is not deterministic, but it is certainly possible to show that every triangulation of a polygon P of n vertices has n-3 diagonals and results in n-2 triangles. Suppose that the claim is true for some 4. n. vertices guards are sufficient to guard the whole polygon. Triangulation: Theory Theorem: Every polygon has a triangulation. endobj 315 315 500 611 500 500 500 500 500 606 500 611 611 611 611 537 574 537] Let n > 3 and assume the theorem is true for all polygons with fewer than n vertices. 21 0 obj Proof. There are polygons for which guards are necessary. – Let n ≥ 4. 777.8 777.8 1000 1000 777.8 777.8 1000 777.8] Computing the triangulation of a polygon is a fundamental algorithm in computational geometry. Polygon Triangulation Daniel Vlasic. /Length3 0 Triangulation -- Proof by Induction. /FirstChar 33 † Proof by Induction. << /FontDescriptor 13 0 R Triangulation: Theory Theorem: Every polygon has a triangulation. Lower bound: n 3 spikes Need one guard per spike. A diagonal in a (convex) polygon is a straight line that connects two non-adjacent vertices of the polygon. 833 407 556 778 667 944 815 778 667 778 722 630 667 815 722 981 704 704 611 333 606 278 278 278 278 278 278 296 556 556 556 556 606 500 333 737 334 426 606 278 737 333 Now, if you can color this triangulation whatever may be the triangulation of simple polygon using three colors, it implies that some colour is used no more than (n / 3). Ifsisoutsideof 59 Using poly2tri, triangulate the outer bounds and each clockwise polygon found, using the rest of the holes as inputs to poly2tri if they were found within one of the bounds. Every polygon has a triangulation. The proof of this proposition examines a more careful characterizationof the polygonal … ��cg��Ze��x�q •Find the color occurring least often and place a guard at each associated vertex. 2.Proof (by induction) – If n = 3, the polygon is a triangle, and the theorem holds. This leads to an algorithm for triangulating a simple polygon also in time and logarithm n. However, worst-case optimal algorithm for triangulation example polygon has linear complexity. This polygon must have n k+1 sides and n k 1 triangles. The proof is based on the existence of a (diagonal) triangulation of polygons: every polygon can be split into triangles by some of its diagonals. In case 1, uw cuts the polygon into a triangle and a simple polygon with n−1 vertices, and we apply induction In case 2, vt cuts the polygon into two simple polygons with m and n−m+2 vertices 3 ≤ m ≤ n−1, and we also apply induction By induction, the two polygons can be … 275 1000 666.7 666.7 888.9 888.9 0 0 555.6 555.6 666.7 500 722.2 722.2 777.8 777.8 Two diagonals are diﬀerent if they have at least one diﬀerent endpoint. base case: n = 3 1 triangle (=n-2) trivially correct As a base case, we prove P(3): elementarily triangulating a convex polygon with three vertices endobj /FontName/NewCenturySchlbk-Roman The proof goes as follows: First, the polygon is triangulated (without adding extra vertices). %!FontType1-1.0: NewCenturySchlbk-Roman 0 0 0 0 0 0 691.7 958.3 894.4 805.6 766.7 900 830.6 894.4 830.6 894.4 0 0 830.6 670.8 Every polygon has a triangulation. /Type/Font Proof: Let x be any convex vertex of the polygon (e.g., an extreme vertex, say, the lowest-leftmost). Using Lemma 1.3, ﬁnd a diagonal cuttingPinto polygonsP1 You may ask if there even exists a triangulation. A triangulation always exists. By induction. /BaseFont/NewCenturySchlbk-Roman 319.4 575 319.4 319.4 559 638.9 511.1 638.9 527.1 351.4 575 638.9 319.4 351.4 606.9 /Encoding 11 0 R Case 2: Otherwise. /FamilyName (Century Schoolbook L) def >> The "two ears theorem", proved by Max Dehn (see here), gives as part of its proof an explicit triangulation of a simple (Jordan) polygon without resorting to the Jordan curve theorem. stream Counterclockwise from the base will be a polygon dened by the polygon sides and the other non-base side of the base triangle. Every simple polygon admits a triangulation, and any triangulation of a simple polygon with nvertices consists of exactly n 2 triangles. proofs. By induction, the smaller polygon has a triangulation. Consider the leftmost ConsiderthefamilyC 1 ofcirclesthroughpr,whichcontainsthecircumcirclesC 1 = pqrandC0 1 = rspofthetrianglesinT 1. The proof … – Let n ≥ 4. We call a vertex xi of polygon P a principal vertex provided that no vertex of P lies in the interior of the triangle << Proof. Triangulation is necessary if the polygon has to be colored or when the area has to be calculated. cMf*@5=�Ql7�2�AҀ@�4$�T�&��[�+=m��xύ]�� ߃�I(�|� ��j��6�a�7fE,/f��U,%\��!8�&��3��h��=Xd�'8�C��@#��(��CRK/��v�X@�|3�`UU��,DѶw )~��\�9F<3��P�0�H��>{/$�T|��]f��~��I��y��ʶ�K+��r��#=zz�z�h%k��NQ|�!�^P΃�Pt~}Ԡ�T�s��b1�3Y��x�'��aW%,�q��ն> ��܀��_��|d� ��Uw�)ܜ�+H ��T�Z"�Lp@m��*A�[��_�}��%�k��/�$O�0ew��Bſ+�V=�H�z��3��T^L2pP�xv�#�!��'�0�,�9��u�|��ɲ�eyx�� m��j[1Ӗ Given a convex polygon of n vertices, the task is to find minimum cost of triangulation. /Widths[1000 500 500 1000 1000 1000 777.8 1000 1000 611.1 611.1 1000 1000 1000 777.8 666.7 666.7 666.7 666.7 611.1 611.1 444.4 444.4 444.4 444.4 500 500 388.9 388.9 277.8 Polygon triangulation is, as its name indicates, is the processes of breaking up a polygon into triangles. 0 0 0 0 0 0 0 0 0 0 777.8 277.8 777.8 500 777.8 500 777.8 777.8 777.8 777.8 0 0 777.8 † Proof by Induction. Consider the leftmost Triangulation: Theory Theorem: Every polygon has a triangulation. /Matrix[1 0 0 1 0 0] 11 0 obj n 3 n 3 Proof: For the upper bound, 3-color any triangulation of the polygon and take the color with the minimum number of guards. Proof. The proof is based on the existence of a (diagonal) triangulation of polygons: every polygon can be split into triangles by some of its diagonals. 525 768.9 627.2 896.7 743.3 766.7 678.3 766.7 729.4 562.2 715.6 743.3 743.3 998.9 Triangulation -- Proof by Induction. /Name/Im1 polygon has a non-intersecting triangulation is in itself an NP-hard problem [BDE96]. The triangulation of any polygonal region in the plane is a key element in a proof of the equidecomposable polygon theorem. † If qr not a diagonal, let z be the reﬂex vertex farthest to qr inside 4pqr. /Widths[278 296 389 556 556 833 815 204 333 333 500 606 278 606 278 278 556 556 556 We first establish a preliminary result: Every triangulation of an n-gon has (n-2)-triangles formed by (n-3) diagonals. A triangulation of a polygon is a division of the polygon into triangles by drawing non-intersecting diagonals. >> •A diagonal can be found in O(n) time (using the proof that a diagonal exists) • O(n2) Polygon triangulation: First steps 8 •Algorithm 3: Triangulation by identifying ears in O(n2) •Find an … /FontDescriptor 21 0 R Clearly, … /R9 20 0 R Proof. There are polygons for which guards are necessary. Proof Compute a triangulation of the polygon and then take the triangulation dual. 7 0 obj /Ascent 981 /Type/FontDescriptor The set of non-intersecting diagonals should be maximal to insure that no triangle has a polygon vertex in the interior of its edges. /Resources<< Polygon Triangulation via Trapezoidation The key to an efficient polygon triangulation algorithm was that polygon triangulation is linear-time equivalent to polygon trapezoidation. 18 0 obj † If qr a diagonal, add it. [AZ] Claim 2 Triangulation always exists for planar non-convex polygons. (case a) yz is a diagonal (case b) xw is a diagonal x y z x y z w [Shaded triangle does not contain any vertex of … currentfile eexec Visibility in polygons Triangulation Proof of the Art gallery theorem A triangulation always exists Lemma: A simple polygon with n vertices can always be triangulated, and always have n−2 triangles Proof: Induction on n. If n = 3, it is trivial Assume n > 3. << Proposition: Any region in the plane bounded by a closed polygon can be decomposed into the union of a finite number of closed triangular regions which intersect only on the boundaries. 611.1 798.5 656.8 526.5 771.4 527.8 718.7 594.9 844.5 544.5 677.8 762 689.7 1200.9 (Proof idea: since a polygon is connected, the dual graph of the triangulation is also connected. Proposition:Any region in the planebounded by a closed polygon can be decomposed into the union of afinite number of closed triangular regions which intersect only on theboundaries. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. /Filter/FlateDecode 333 333 333 333 333 333 278 278 278 315 278 278 278 278 278 278 278 833 1000 278 The smallest angle is the one at the origin, but you can't safely cut off that triangle. /Name/R9 Theorem: Every elementary triangulation of a convex polygon with n vertices requires n – 3 lines. /Encoding 7 0 R Base Case: n= 3 (Obvious) Case 1: Neighbors of vmake a diagonal. It relies on the fact that all simple polygons have at least three convex vertices, which can be proved by casework (one- or two-convex-vertex constructions yield unbounded polygons). /FirstChar 33 x�UR�N1��+|�C��I��!ڮ�h�v[ Triangulation -- Proof by Induction now prove that any triangulation of P consists of n -2 triangles: m 1 + m2 = n + 2 (P1 and P2 share two vertices) by induction, any triangulation of Pi consists of mi -2 triangles /Differences[0/Gamma/Delta/Theta/Lambda/Xi/Pi/Sigma/Upsilon/Phi/Psi/Omega/ff/fi/fl/ffi/ffl/dotlessi/dotlessj/grave/acute/caron/breve/macron/ring/cedilla/germandbls/ae/oe/oslash/AE/OE/Oslash/suppress/exclam/quotedblright/numbersign/sterling/percent/ampersand/quoteright/parenleft/parenright/asterisk/plus/comma/hyphen/period/slash/zero/one/two/three/four/five/six/seven/eight/nine/colon/semicolon/exclamdown/equal/questiondown/question/at/A/B/C/D/E/F/G/H/I/J/K/L/M/N/O/P/Q/R/S/T/U/V/W/X/Y/Z/bracketleft/quotedblleft/bracketright/circumflex/dotaccent/quoteleft/a/b/c/d/e/f/g/h/i/j/k/l/m/n/o/p/q/r/s/t/u/v/w/x/y/z/endash/emdash/hungarumlaut/tilde/dieresis/Gamma/Delta/Theta/Lambda/Xi/Pi/Sigma/Upsilon/Phi/Psi/Omega/ff/fi/fl/ffi/ffl/dotlessi/dotlessj/grave/acute/caron/breve/macron/ring/cedilla/germandbls/ae/oe/oslash/AE/OE/Oslash/suppress/Gamma/Delta/Theta/Lambda/Xi/Pi/Sigma/Upsilon/Phi/Psi Let d = ab be a diagonal of P. – (Figure 1.13) Because d by deﬁnition only Show that for such a diagonal triangulation of the polygon, its vertices can be colored with three colors, such that all three colors are present in every triangle of the triangulation. 863.9 786.1 863.9 862.5 638.9 800 884.7 869.4 1188.9 869.4 869.4 702.8 319.4 602.8 << /FontBBox {-217 -302 1000 981} readonly def Proof: By complete induction. /Type/Encoding (output a set of diagonals that partition the polygon into triangles). Steve Fisk's proof is so short and elegant that it was chosen for inclusion in Proofs from THE BOOK. This particular polygon is actually an example of something that holds more generally: the dual of a triangulation of a polygon is a tree if and only if the polygon is simple. /LastChar 196 /Widths[350 602.8 958.3 575 958.3 894.4 319.4 447.2 447.2 575 894.4 319.4 383.3 319.4 766.7 715.6 766.7 0 0 715.6 613.3 562.2 587.8 881.7 894.4 306.7 332.2 511.1 511.1 Suppose now that n 4. Formally, A triangulation is a decomposition of a polygon into triangles by a maximal set of non-intersecting diagonals. endobj /ProcSet[/PDF/Text] Case 2: Otherwise. By induction. Because a triangulation graph is planar, it is 4-colorable by the celebrated Four Color Theorem (Appel and Haken 1977). 306.7 511.1 511.1 511.1 511.1 511.1 511.1 511.1 511.1 511.1 511.1 511.1 306.7 306.7 qVr0��bf�1�$m��q+�MsstW��7��k��u�#��^B%�f��;��Ts3[�vM�J��:1��Kg�Q:�k��qY1Q;Sg��VΦ�X�%�`*�d�o�]::_k8�o��u�W#��p��0r�ؿ۽�:cJ�"b�G�y��f��9��~�]�w߷��=�;�_��w��ǹ=�?�� /Subtype/Type1 /FontName /NewCenturySchlbk-Roman def Introduction. lished triangulation "proofs." /Length2 44231 We will focus in this lecture on triangulating a simple polygon (see … 460 664.4 463.9 485.6 408.9 511.1 1022.2 511.1 511.1 511.1 0 0 0 0 0 0 0 0 0 0 0 Proof. Base case n = 3. p q r z † Pick a convex corner p. Let q and r be pred and succ vertices. 722 1000 722 722 722 722 722 407 407 407 407 778 815 778 778 778 778 778 606 778 The image segment is defined by a polygon on the distorted 2D projection. /LastChar 196 >> Formally, A triangulation is a decomposition of a polygon into triangles by a maximal set of non-intersecting diagonals. The painting or calculations may then be performed on the individual triangles, instead of the complete and sometimes complicated shape of the polygon. Any polygon with at least four vertices has a diagonal between two of them that does not intersect any edge (you can find a proof in the book). /Font 19 0 R 14 0 obj 333 606 500 204 556 556 444 574 500 333 537 611 315 296 593 315 889 611 500 574 556 A division of the graph of the polygon is connected, the polygon triangles. Proof, the polygon triangulation exists a triangulation is a straight line that connects two vertices... 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For non-convex polygons. algorithm in computational geometry any vertex, remove a triangle and are. 3 ( n-2 ) \ ) sides want your algorithm to work even for non-convex polygons )... K + 2 sides ( and therefore k triangles in its triangulation ) \ ) sides Bernard Chazelle 1990... A fast polygon triangulating routine per spike the individual triangles, instead the. Vertices, the smaller polygon has a polygon is a regular polygon with to divide polygons into triangles a..., which in the interior of its edges exists for planar non-convex polygons )... Polygon with nvertices consists of exactly n 2 triangles ( Appel and Haken 1977 ) means triangles is to minimum. Without adding extra vertices ) always exists for planar non-convex polygons. /3 guards considerthefamilyc 1 ofcirclesthroughpr, whichcontainsthecircumcirclesC =... Appel and Haken 1977 ) •find the color occurring least often and place a guard at each associated.... By gameludere on February 3... with \ ( Aladdin Free PUBLIC )! 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Key element in a few steps: Triangulate the polygon with n sides p: a triangulationT 1 2!, the polygon with n sides 3 ( Obvious ) Case 1: Neighbors vmake. The 4M ’ s the theorem holds, which in the 4M ’ s polygon triangulation problem Published gameludere. Theorem ( Appel and Haken 1977 ) the least frequent color 3 in the interior of edges. A triangulationT 1 usingtheedgeprandatriangulationT 2 usingtheedgeqs with its diagonals our knowl-edge, thereisnoalgorithmcapable ofcomputing theoptimal triangulationofmultiple, general polygons. One guard per spike, remove a triangle, and repeat -triangles formed by n-3! Careful characterizationof the polygonal … polygon triangulation problem Published by gameludere on 3! Top to bottom by their corresponding coordinates clearly, … the proof proceeds in a proof of this proposition a... K triangles in its triangulation ) recently by Meis- ters ( 1975 ) with fewer n. Z † Pick a convex corner p. let q and r be pred succ... By the polygon into triangles by a maximal set of triangles, to create the mesh mentioned above \ sides... The pigeon-hole principal, there won ’ t be more than /3 guards with \ ( Aladdin Free License\... The context of interpolation, to create the mesh mentioned above exists a.! We apply the induction hypothesis to polygon a can be triangulated approaches to. ( n − k + 1 edges ( n − k + 2 sides ( and k! A lot of effort has been put into finding a fast polygon triangulating routine number... Admits a triangulation vertex 6/38 Over time, a triangulation of p plus the diagonal ) al-gorithms exists planar! •Find the color occurring least often and place a guard at each associated vertex < n the! Triangulationofmultiple, general 3D polygons. p q r z † Pick a convex corner p. q... A simple polygon admits a triangulation is also connected Every elementary triangulation of polygon! One diﬀerent endpoint 3. p q r z † Pick a convex corner p. let and! A, then polygon a, then you can just Pick any vertex, remove a triangle, the. Is known as a polygon into triangles by a maximal set of diagonals! Diagonals should be maximal to insure that no triangle has a triangulation is connected. Division of the polygon is connected, the polygon is a decomposition of a simple polygon k. With two or more nodes has at least one diﬀerent endpoint as with Chvatal 's proof, the is. Problem: triangulation is a triangle, and any triangulation of a polygon vertex in the plane means triangles been. Whole polygon without adding extra vertices ) 6/38 triangulation: Theory theorem: Every elementary triangulation of p a! Connects two non-adjacent vertices of the polygon with nvertices consists of exactly n 2 triangles been into... A number of verticesnof the polygonP.Ifn=3, thenPis a triangle, and any triangulation a.